这是代码正在运行的代码,我希望输出作为降序计数,如果计数是相同的,则按名称订购。
from collections import Counter
import re
from nltk.corpus import stopwords
import operator
text = "The quick brown fox jumped over the lazy dogs bowl. The dog was angry with the fox considering him lazy."
def tokenize(text):
tokens = re.findall(r"w+|S", text.lower())
#print(tokens)
tokens1 = []
for i in tokens:
x = re.findall(r"w+|S", i, re.ASCII)
for j in x:
tokens1.append(j)
return tokens
tok = tokenize(text)
punctuations = ['(',')',';',':','[',']',',', '...', '.', '&']
keywords = [word for word in tok if not word in punctuations]
cnt = Counter()
d= {}
for word in keywords:
cnt[word] += 1
print(cnt)
freq = operator.itemgetter(1)
for k, v in sorted(cnt.items(), reverse=True, key=freq):
print("%3d %s" % (v, k))
当前输出:
4 the
2 fox
2 lazy
1 quick
1 brown
1 jumped
1 over
1 dogs
1 bowl
1 dog
1 was
1 angry
1 with
1 considering
1 him
所需的输出:
4 the
2 fox
2 lazy
1 angry
1 bowl
1 brown
1 considering
1 dog
1 dogs
等。
使用返回元组的排序功能。元组中的第一个项目是计数的倒数(字典中的值(,第二个是字符串(字典中的键(。您可以通过删除变量freq
来做到这一点,从而在呼叫中删除关键字reverse
,并为字典中的每个项目返回(-Value,key(返回(-value,键(的少量lambda函数。该程序的最后几行是:
print(cnt)
for k, v in sorted(cnt.items(), key=lambda item: (-item[1], item[0])):
print("%3d %s" % (v, k))
lambda函数中的符号的原因是获得正确的排序顺序,因为默认排序顺序最低至最高。